What are the solutions to the equation (n-2)(n+4)=27 ?

The solutions to the equation (n-2)(n+4)=27 are n=5,n=-7

Find the zeros of (n-2)(n+4)=27

The zeros of (n-2)(n+4)=27 are n=5,n=-7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(n - 2) (n + 4) = 27

n = 5, n = - 7

Solution steps

(n - 2) (n + 4) = 27

Expand

(n - 2) (n + 4) : n^{2} + 2 n - 8

n^{2} + 2 n - 8 = 27

Move

27

to the left side

n^{2} + 2 n - 35 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 35 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 35) = 12

n_{1, 2} = \frac{- 2 \pm 12}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 2 + 12}{2 \cdot 1}, n_{2} = \frac{- 2 - 12}{2 \cdot 1}

n = \frac{- 2 + 12}{2 \cdot 1} : 5

n = \frac{- 2 - 12}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

n = 5, n = - 7

0.44048 x^{2} - 60.726 x + 1848.6 = 0

326 = x^{2}

a^{2} - 6 a + 9 = 0

n^{2} + 6 n + 2 = 0

z^{2} + 25 = 0

What are the solutions to the equation (n-2)(n+4)=27 ?
The solutions to the equation (n-2)(n+4)=27 are n=5,n=-7
Find the zeros of (n-2)(n+4)=27
The zeros of (n-2)(n+4)=27 are n=5,n=-7